 | Jeremiah Day - 1815 - 388 páginas
...solidity of a frustum of a triangular pyramid is equal to -5 of the height, multiplied into the »um of the areas of the two ends and the square root of the product of these areas. This is true also of a frustum of any other pyramid. (Sup. Euc. 12, 3. Cor. 2.) If the smaller end... | |
 | 1818 - 264 páginas
...portion of the upper end is cut off, by a plane paralell to the base. RULE. Add together the ureas of the two ends, and the square root of the product...multiply the sum by £ of the perpendicular height, and the result will be the sojid content. OR 2. Divide the difference of the cubes of the diameters... | |
 | Jeremiah Day - 1824 - 440 páginas
...and 44, what is the whole surface ? Half the difference of the diameters is 18. And ^18" +243 =30, the slant-height, (Art. 52.) The convex surface of...AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS J AND MULTIPLY THE. SUM BV | OF THE PERPENDICULAR HEIGHT. This rule, which was given for the frustum... | |
 | John Bonnycastle - 1829 - 256 páginas
...last product will be the solidity. Note. — When the ends of the pyramids are not regular polygons. Add together the areas of the two ends and the square root of their product; multiply the sum by the height, and 1 of the product will be the solidity. EXAMPLES.... | |
 | Jeremiah Day - 1831 - 418 páginas
...the two ends 80 and 44, what is the whole surface ? Half the difference of the diameters is 1 8. 30, the slant-height, (Art. 52.) The convex surface of...AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS J AND MULTIPLY THE SUM BY \ OF THE PERPENDICULAR HEIGHT. Ex. 1. What is the solidity of a mast which... | |
 | Jeremiah Day - 1831 - 394 páginas
...unequal triangles which form its sides. PROBLEM V. To fold the SOLIDITY of a FRUSTUM of a pyramid. 50. ADD TOGETHER THE AREAS OF THE TWO ENDS, AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS J AND MULTIPLY THE SUM BY i OF THE PERPENDICULAR HEIGHT OF THE SOLID. Let CDGL (Fig. 17.) be a vertical... | |
 | Ira Wanzer - 1831 - 408 páginas
...PROBLEM III. — To find the anlid content of any Frustum of a Pyramid. RULE I. — Add into one sum, the areas of the two ends and the square root of the product of those areas, and onethird of this sum will be the mean area of a section between the two bases ; which... | |
 | Jeremiah Day - 1831 - 520 páginas
...surface is 1 2390 PROBLEM VI. To faid the SOLIDITY of a FRUSTUM of a cone. 68. ADD TOGETHER THE AREAS OP THE TWO ENDS, AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS J AND MULTIPLY THE SUM BY £ OF THE PERPENDICULAR HEIGHT. Ex. 1. What is the solidity of a mast which... | |
 | Jeremiah Day - 1836 - 418 páginas
...The sum of the areas of the two ends is 6547 And the whole surface is 12390 X PROBLEM VI. To find die SOLIDITY of a FRUSTUM of a cone. 68. ADD TOGETHER...THE PRODUCT OF THESE AREAS; AND MULTIPLY THE SUM BY 1 OF THE PERPF.NDICULAR HEIGHT. This rule, which was given for the frustum of a pyramid, (Art. 50.)... | |
 | 1836 - 488 páginas
...slant hight, by the sum of the peripheries of the two ends. Toßnd the solidity of afrustrum of a cone. Add together the areas of the two ends, and the square...root of the product of these areas ; and multiply the same by £ of the perpendicular hight. Toßnd the surface of a sphere. Multiply the diameter by the... | |
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PROBLEM VI. To find the SOLIDITY of a FRUSTUM of a cone. 68. ADD TOGETHER THE AREAS OF THE TWO ENDS, AND THE SQUARE ROOT OF THE PRODUCT OF THESE AREAS; AND MULTIPLY THE SUM BY 1 OF THE PERPENDICULAR HEIGHT. 
